A) A hyperbola with the length of conjugate axis 3
B) a hyperbola with eccentricity\[\sqrt{5}\]
C) an ellipse with the length of major axis 6
D) an ellipse with eccentricity\[\frac{2}{\sqrt{5}}\]
Correct Answer: A
Solution :
\[tx-2y-3t=0\] \[x-2ty+3=0\] \[\frac{\begin{align} & tx-2y-3t=0 \\ & tx-2{{t}^{2}}y\underline{+}3t=0 \\ \end{align}}{y(2{{t}^{2}}-2)=6t}\] \[\frac{\begin{align} & {{t}^{2}}x-2ty-3{{t}^{2}}=0 \\ & -x-2ty\underline{+}3=0 \\ \end{align}}{({{t}^{2}}-1)x=(3{{t}^{2}}+1)}\] \[y\frac{6t}{2{{t}^{2}}-2}=\frac{3t}{{{t}^{2}}-1}\] \[x=-3\sec 2\theta \] \[2y=3(-tan2\theta )\] \[{{\sec }^{2}}2\theta -{{\tan }^{2}}2\theta =1\] \[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{9/4}=1\] \[{{a}^{2}}=9;\] \[{{b}^{2}}=9/4\] \[\lambda (T.A)=6\] \[;{{e}^{2}}=1+\frac{9/4}{9}=1+\frac{1}{4};e\,e=\frac{\sqrt{5}}{2}\]
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